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Step 2: Calculate the Expected Value of \(N#\)tag_content#Now that we have the probability \(P_N(n)\) for each value of \(n\), we can calculate the expected value of \(N\), \(E[N]\). The expected value can be computed as the sum of the product of each possible value of \(n\) and its corresponding probability \(P_N(n)\): $$ E[N] = \sum_{n=1}^{\infty} n \cdot P_N(n) $$ We now need to calculate the sum of this infinite series. To do this, we can compute the first few probabilities and recognize a pattern that generalizes to the sum: $$ \begin{aligned} E[N] &= 1 \cdot P_N(1) + 2 \cdot P_N(2) + 3 \cdot P_N(3) + \cdots \\ &= 1 \cdot 0 + 2 \cdot \frac{1}{2!} + 3 \cdot \left(-\frac{1}{3!}\right) + 4 \cdot \frac{2}{4!} - \cdots \\ &= 1 \cdot 0 + \left(\frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \cdots\right) + 2 \cdot \left(-\frac{1}{6} + \frac{1}{24} - \cdots\right) + \cdots \\ \end{aligned} $$ Now, let's examine the expressions in the parenthesis. The expression in the first parenthesis is an alternating geometric series with a ratio \(-1/3\) and the sum can be found as, $$ \frac{1/2}{1-(-1/3)}=\frac{3}{5} $$ Similarly for the second parenthesis, we get $$ -\frac{1}{6}\left(\frac{1}{1-(-1/3)}\right)=-\frac{1}{10} $$ And the rest of the series follows a similar pattern. We can observe that the general pattern that emerges is: $$ E[N] = \frac{3}{5} - \frac{1}{10} + \cdots = \sum_{n=1}^{\infty} (-1)^{n+1}\cdot\frac{n}{(n+1)(n+2)} $$ Summing up this series, we get $$ E[N] = \sum_{n=1}^{\infty} (-1)^{n+1}\cdot\frac{n}{(n+1)(n+2)} = \frac{1}{2} - \frac{1}{6} + \frac{1}{12} - \cdots = \boxed{\frac{1}{2}} $$ So, the expected value of \(N\) is \(\frac{1}{2}\).